# Let $y=y=\sqrt{e^\sqrt x },x>0\;find\;\large\frac{dy}{dx}.$

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• Use chain rule.
$\large\frac{dy}{dx}\: \large\frac{1}{2\sqrt{e^{\sqrt x}}}$ x $e^{\sqrt x}$ x $\large\frac{1}{2\sqrt x}= \large\frac{\sqrt{e^{\sqrt x}}}{4\sqrt x}$

edited Mar 26, 2013